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Question: How do you write \(y = - 10{x^2} + 2\) in vertex form?...

How do you write y=−10x2+2y = - 10{x^2} + 2 in vertex form?

Explanation

Solution

We have to rewrite the given equation using some algebraic steps in order to make it look like the Standard Vertex form of the quadratic function.
For this the first step is to isolate the x2{x^2} terms by moving 22 to the other side of the equal sign.
Then, we need a leading coefficient of 11 for completing the square so factor out the current leading coefficient of −10 - 10. Next, we have to create a perfect square.
Isolate the yy-term. So move 22 to the other side of the equal sign. The equation obtained will be the vertex form of the given equation.

Formula used: Vertex form of the quadratic function:
y=a(x−h)2+ky = a{\left( {x - h} \right)^2} + k where (h,k)\left( {h,k} \right) is the vertex or the “center” of the quadratic function or the parabola.

Complete step-by-step solution:
Given quadratic function: y=−10x2+2y = - 10{x^2} + 2
This quadratic equation is in the formy=ax2+bx+cy = a{x^2} + bx + c.
However, we need to rewrite it using some algebraic steps in order to make it look like this
y=a(x−h)2+ky = a{\left( {x - h} \right)^2} + k
This is the vertex form of the quadratic function where (h,k)\left( {h,k} \right) is the vertex or the “center” of the quadratic function or the parabola.
Here, the coefficient ofx2{x^2},a=−10a = - 10 not equal to11.
Since we will be completing the square, we will isolate the x2{x^2} terms.
So, the first step is to move 22 to the other side of the equal sign.
Subtract 22 from both sides of the equation y=−10x2+2y = - 10{x^2} + 2.
⇒y−2=−10x2\Rightarrow y - 2 = - 10{x^2}
Now, we need a leading coefficient of 11 for completing the square so factor out the current leading coefficient of −10 - 10.
⇒y−2=−10(x2)\Rightarrow y - 2 = - 10\left( {{x^2}} \right)
Now, we have to create a perfect square.
⇒y−2=−10(x−0)2\Rightarrow y - 2 = - 10{\left( {x - 0} \right)^2}
Isolate the yy-term. So move 22 to the other side of the equal sign.
Add 22 to both sides of the equation.
⇒y=−10(x−0)2+2\Rightarrow y = - 10{\left( {x - 0} \right)^2} + 2

Therefore, the given equation in vertex form is y=−10(x−0)2+2y = - 10{\left( {x - 0} \right)^2} + 2.

Note: We can also solve this type of question using Sneaky Tidbit Method.
When working with the vertex form of a quadratic function: h=−b2ah = \dfrac{{ - b}}{{2a}} and k=f(h)k = f\left( h \right)
The “aa” and “bb” referenced here refers to f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c.
Compare y=−10x2+2y = - 10{x^2} + 2 with f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c and determine the value of aa, bb and cc.
Here, y=−10x2+2y = - 10{x^2} + 2 can be written as y=−10x2+0x+2y = - 10{x^2} + 0x + 2.
So, a=−10a = - 10, b=0b = 0 and c=2c = 2.
Now, find the value of hh by putting the values of aa and bb in h=−b2ah = \dfrac{{ - b}}{{2a}}.
Here, a=−10a = - 10 and b=0b = 0.
So, h=−02×(−10)h = \dfrac{{ - 0}}{{2 \times \left( { - 10} \right)}}
On simplification, we get
⇒h=0\Rightarrow h = 0
Now, find kk by putting the value of hh in f(x)f\left( x \right).
⇒k=f(0)\Rightarrow k = f\left( 0 \right)
Put f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c in the above equation.
⇒k=a(0)2+b(0)+c\Rightarrow k = a{\left( 0 \right)^2} + b\left( 0 \right) + c
Put the values of aa, bb and cc in the above equation.
Here, a=−10a = - 10, b=0b = 0 and c=2c = 2.
⇒k=(−10)(0)2+0(0)+2\Rightarrow k = \left( { - 10} \right){\left( 0 \right)^2} + 0\left( 0 \right) + 2
On simplification, we get
k=2k = 2
Therefore, the vertex is (h,k)=(0,2)\left( {h,k} \right) = \left( {0,2} \right).
Now, put the value ofaa, hh and kk in Standard Vertex form of the quadratic function.
Since, Vertex form of the quadratic function:
y=a(x−h)2+ky = a{\left( {x - h} \right)^2} + k where (h,k)\left( {h,k} \right) is the vertex or the “center” of the quadratic function or the parabola.
Here, a=−10a = - 10, h=0h = 0 and k=2k = 2.
Therefore, the given equation in vertex form is y=−10(x−0)2+2y = - 10{\left( {x - 0} \right)^2} + 2.